[Leetcode]Maximum Subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

class Solution {public:    /*algorithm  divide and conquer      A[1..n] can divide to A[1..i-1] A[i] A[i+1..n]      f(1..n) = max(f(1..i-1),f(i+1...n),union(f(1..i-1)+A[i]+f(i+1..n))      time O(nlogn) space O(nlogn)    */    //[start,end)    int maxSubArrayHelper(vector<int>&nums,int start,int end)    {        if((end - start) < 1)return INT_MIN;        if((end-start) < 2)return nums[start];        int mid = start + (end-start)/2;        int left = maxSubArrayHelper(nums,start,mid);        int right = maxSubArrayHelper(nums,mid+1,end);        int lsum = 0,lsmax=0;        for(int k=mid-1;k >= start;k--){            lsum += nums[k];            lsmax = max(lsmax,lsum);        }        int rsum = 0,rsmax=0;        for(int k=mid+1;k < end;k++){            rsum += nums[k];            rsmax = max(rsum,rsmax);        }        return max(max(left,right),lsmax+nums[mid]+rsmax);    }    int maxSubArray(vector<int>& nums) {        return maxSubArrayHelper(nums,0,nums.size());    }};

class Solution {public:    /*algorithm dp solution     for array,A[1...n]     f(1)=A[1];     f(2)=max(f(1) + A[2],A[2])     f(n) = max(f(n-1) + A[n],A[n])     time O(n) space O(1)    */    int maxSubArray(vector<int>& nums) {        int n = nums.size();        int gmax = nums[0];        int f0 = nums[0],f1;        for(int i = 1;i < n;i++){            f1 = max(f0+nums[i],nums[i]);            f0 = f1;            gmax = max(gmax,f0);        }        return gmax;    }};